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Is there only one reversible way to move from one state to another?

If we consider two states $A$ and $B$ on an isotherm and we move from $A$ to $B$ by first reversible isochoric process and then reversible isobaric process. Now the path followed should be reversible since both the processes were reversible. But what about simply following the reversible isothermal process?

According to me both processes should be reversible. Now entropy is the heat added reversibly to move from one state to another divided by the temperature at which it is added. But we know that the heat added to the system is different in both the cases. Then how is entropy a state function?

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    $\begingroup$ Entropy change is not heat added devided by the temperature in general .In general it is $\Delta S=\int_{\mathrm{rev}}\frac{dQ}{T},$ . $\endgroup$ – Paul Mar 16 '17 at 17:42
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The total heat added in both the processes is different. Change in entropy is defined as $\int(dQ/T)$. Along the isotherm, the temperature remains constant. But along the other two reversible processes you have mentioned, the temperature is not constant. Effectively, it can be seen by integration that change in entropy in both processes is the same.

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Your question goes right in the kernel of the meaning of the term state function.

A state function is a function defined over all possible states of the system such that its value for every state does not depend on how the system reached the state. Each state has a definite and unique value for the given state function.

The state $A$ has a definite value for the state function entropy, $S(A)$. The same for the state $B$, which gives $S(B)$. Thus the difference in entropy between the states $A$ and $B$ is simply $\Delta S=S(B)-S(A)$ and this value does not depend on the process that takes $A$ to $B$. The difference $\Delta S$ between $A$ and $B$ exists even for irreversible paths and it has always the same value.

In the case of entropy, there is some subtlety though. The way we calculate the difference $\Delta S$ is always $$\Delta S=\int_{\mathrm{rev}}\frac{dQ}{T},$$ where the integral has to be computed through a reversible process. There is a plenty of reversible process from $A$ to $B$ but we just choose the simplest one for calculations.

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There are generally many reversible paths between different states. As an example consider a Carnot cycle. If I start at the beginning of the cycle at the point where the working medium is fully contracted I can, since every step is reversible, get to the point half way through, where the working medium is fully expanded, by going around the cycle in either direction, giving me two different paths between the two states.

Notice however that both these paths have two stages. On one you expand isothermally and then adiabatically, and on the other you apply the same steps in the reverse order.

There are, however, limits on which states can be accessed on certain types of path. In particular you can only move between two states adiabatically if they lie on the same adiabat and can only move between them isothermally if they lie on the same isotherm.

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  • $\begingroup$ What about entropy...how is it a state function then? $\endgroup$ – oshhh Mar 16 '17 at 16:34
  • $\begingroup$ @By Symmetry, could you please explain the reason for this statement? "There are, however, limits on which states can be accessed on certain types of path". Thank you. $\endgroup$ – Tejas P Mar 16 '17 at 16:35
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    $\begingroup$ @OsheenSachdev Clausius Theorem, when applied to the reversible case, tells you that the change in entropy along any reversible path is the same, and so entropy is a well defined state function. I agree that this is a surprising given that we have integrated something which is not a state function along an arbitrary path and somehow have end up with something path independent. The proof can be found in any thermodynamics textbook $\endgroup$ – By Symmetry Mar 16 '17 at 16:46
  • $\begingroup$ @TEJASP It would have been clearer if I had said, given a certain initial state, there are limits on which states can be accessed. So if you picture states on a $PV$ diagram, then, for an ideal gas, the only states that I can access along an isothermal path are those the satisfy $PV = const$ whereas for an adiabatic path states must lie on the curve $PV^\gamma = const$. $\endgroup$ – By Symmetry Mar 16 '17 at 16:52
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Entropy is surely a state function which only depends on your start and end states, and the change in entropy between two states is defined by integrating infinitesimal change in entropy along a reversible path. But heat $Q$ is not a state variable, the amount of heat gained or lost is path-dependent. Once we divided $\mathrm{d}Q$ with temperature, this gives a exact differential $\mathrm{d}S$. Therefore, temperature is often mathematically referred to as the "integrating factor" (see https://en.wikipedia.org/wiki/Integrating_factor) of $\mathrm{d}Q$.

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